 Solution

For each $n\in \mathbb{N}$, the open intervalmeets $\mathbb{N}$ only in $\{n\}$. Thus every singleton is open in the subspace topology on $\mathbb{N}$. Every subset $A\subseteq \mathbb{N}$ is the union $A={\bigcup }_{n\in A}\{n\}$ and is therefore open. Hence the induced topology is the discrete topology: